Painevg

2021-12-17

Find the points on the given curve where the tangent line is horizontal or vertical.
$r={e}^{\theta }$

encolatgehu

To find the points where the tangents are horizontal/vertical, we need to find dy/dx. Using the chain rule fore differentiation, we can write
$\frac{dy}{dx}=\frac{dy}{d\theta }\cdot \frac{d\theta }{dx}=\frac{\frac{dy}{d}\theta }{\frac{dx}{d}\theta }$
Substitute $y=r\mathrm{sin}\theta ={e}^{\theta }\mathrm{sin}\theta$ and $x=r\mathrm{cos}\theta ={e}^{\theta }$, to get
$\frac{dy}{dx}=\frac{d\frac{{e}^{\theta }\mathrm{sin}\theta }{d}\theta }{d\frac{{e}^{\theta }\mathrm{cos}\theta }{d}\theta }$
Use product rule for differentiation
$\frac{dy}{dx}=\frac{d{\left({e}^{\theta }\right)}^{\prime }\mathrm{sin}\theta +{e}^{\theta }{\left(\mathrm{sin}\theta \right)}^{\prime }}{{\left({e}^{\theta }\right)}^{\prime }\mathrm{cos}\theta +{e}^{\theta }{\left(\mathrm{cos}\theta \right)}^{\prime }}$
$\frac{dy}{dx}=\frac{{e}^{\theta }\mathrm{sin}\theta +{e}^{\theta }\mathrm{cos}\theta }{{e}^{\theta }\mathrm{cos}\theta -{e}^{\theta }\mathrm{sin}\theta }$
Divide both the numerator and the denominator by ${e}^{\theta }$
$\frac{dy}{dx}=\frac{\mathrm{sin}\theta +\mathrm{cos}\theta }{\mathrm{cos}\theta -\mathrm{sin}\theta }$
The tangents will be horizontal when the numerator is 0
$\mathrm{sin}\theta +\mathrm{cos}\theta =0$
Subtract $\mathrm{cos}\theta$ from both sides
$\mathrm{sin}\theta =-\mathrm{cos}\theta$
Divide both sides by $\mathrm{cos}\theta$
$\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=-1$
In the right-hand side, we will rewrite -1 as $\mathrm{tan}\left(-\frac{\pi }{4}\right)$
$tam\theta =\mathrm{tan}\left(-\frac{\pi }{4}\right)$
Remember that: the general solution of the equation $\mathrm{tan}x=\mathrm{tan}\alpha$ is
$x=\alpha +n\pi$
Therefore, the tangents are horizontal when
$\theta =-\frac{\pi }{4}+n\pi$
The tangents will be vertical when the denomiantor is 0
$\mathrm{sin}\theta =\mathrm{cos}\theta$
Divides both sides by $\mathrm{cos}\theta$
$\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=1$
In the right-hand side, we will rewrite 1 as $\mathrm{tan}\left(\frac{\pi }{4}\right)$
$\mathrm{tan}\theta =\mathrm{tan}\left(\frac{\pi }{4}\right)$
Therefore, the tangents are vertical when

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